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Re: PS: Set 7, Q36 - Odd and even numbers [#permalink]
18 Sep 2007, 20:08

gluon wrote:

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer?

A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

Man I hate that the copy pasting drops all formatting!!! I have fixed the question now. Sorry for wasting all your time.

Something isn't right w/ this question. Ignore a/b b/c there are no definite rules about odds and evens for division.

but if a-b is even then a and b are both even or both odd.

A: a can be even so a/2 can be even, not must be odd.
B. same reason as A
C. since a and b can both be even then this doesnt have to be odd
D. a+2/2 a can be even so this doesnt have to be odd.
E. Same reason as D.

a and b both should be even to satusfy the a-b and a/b to be even.
As a/b is even, a>b. The minimum value B can acquire is 2. So a will be greater than 2.

Now close call is between D and E.
a/2 +1 will always be odd as A/2 is even.

b/2 + 1 not neccessarily be odd. For b =2, it's not odd.

Re: PS: Set 7, Q36 - Odd and even numbers [#permalink]
20 Sep 2007, 03:25

gluon wrote:

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer?

A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

Man I hate that the copy pasting drops all formatting!!! I have fixed the question now. Sorry for wasting all your time.

I choose D as well. if a=80 and b=40 then a/2 and b/2 are even. Therefore, A and B are incorrect. If a/2 and b/2 are even, then (a+b)/2 = a/2 + b/2 will be even as well. C is incorrect as well.

I am puzzled because it seems to me that (a+2)/2 = a/2 + 1 will be always odd. The same holds for (b+2)/2. But how can we say that D and E are correct? Am I missing something here?

a and b both should be even to satusfy the a-b and a/b to be even. As a/b is even, a>b. The minimum value B can acquire is 2. So a will be greater than 2.

Now close call is between D and E. a/2 +1 will always be odd as A/2 is even.

b/2 + 1 not neccessarily be odd. For b =2, it's not odd.

vshaunak if b=2 then (a+b)/2 = a/2 + 1, which will be always odd and therefore C gets correct...

Re: PS: Set 7, Q36 - Odd and even numbers [#permalink]
20 Sep 2007, 08:01

gluon wrote:

If a and b are positive integers such that a – b and a/b are both even integers, which of the following must be an odd integer?

A. a/2 B. b/2 C. (a+b)/2 D. (a+2)/2 E. (b+2)/2

Man I hate that the copy pasting drops all formatting!!! I have fixed the question now. Sorry for wasting all your time.

The stem tells us that a-b is even which means that a and b are either both odd or both even.

The stem also tells us that a/b is even (a/b=E). If you multiply b by both sides you see that a =b*E. In this case a has to be even because the product of any number and an even number is even.

To satisfy both requirements a and b must be even.

A. Even/2 could be odd or even (Ex. 8/2 = 4, 14/2=7)
B. same as A
C. same as A
D. This has to be odd becasue a has to be a multiple of 4 as posted earlier.
E. This could both be even or odd as well. Also note b/2 + 1 does NOT have to be odd because a/2 can be odd or even therefore and even + odd = odd and odd + odd = even.

a and b both should be even to satusfy the a-b and a/b to be even. As a/b is even, a>b. The minimum value B can acquire is 2. So a will be greater than 2.

Now close call is between D and E. a/2 +1 will always be odd as A/2 is even.

b/2 + 1 not neccessarily be odd. For b =2, it's not odd.

vshaunak if b=2 then (a+b)/2 = a/2 + 1, which will be always odd and therefore C gets correct...

We have to find the solution, which is always true. B can be 2 or some other even integer. If it's working for b=2 does not necessarily mean that it will work for other values.